Integrating without using the symbolic toolbox

syms x

f=log(45.*sin(x.^2) + cos(x)); % initial function

g=sqrt(1+diff(f).^2); % requirement step 1

h=int(g,0,pi/2) ;% requirement final step

instead of int i want to use " integral " and solve numerically ,if i remove the syms and introduce @(x) then i won't get a displayed result

and the result of final step while using the syms method is :

int(((sin(x) - 90*x*cos(x^2))^2/(45*sin(x^2) + cos(x))^2 + 1)^(1/2), x, 0, pi/2)

x belongs (0,pi/2)

How can i get rid of syms ?

Answer 1

[John D'Errico]

This is an arc length integral. Apparently you wish to compute the length of that curve in the plane, with x going from 0 to pi/2. As a numerical integral, this is easy.

syms x
f=log(45.*sin(x.^2) + cos(x)); % initial function 
fplot(f,[0,pi/2])

g=sqrt(1+diff(f).^2); % requirement step 1
gfun = matlabFunction(g);

First, it alway makes sense to plot everything, before you just throw it into any general tool.

fplot(g,[0,pi/2])

I don't epect that arc length integral has an exact solution. I might also be slightly worried about what seems to be a singularity at 0, though an integration tool should survive that. You can use vpaintegral, if you wish to stay in the symbolic domain.

vpaintegral(g,[0,pi/2])

ans = 

4.73865

Or you can use integral.

integral(gfun,0,pi/2)

ans = 4.7386

Which seems to agree.

And, since I have a tool on the file exchange that can compute the arclength of a curve, I might try this:

ffun = matlabFunction(f);
xint = linspace(0,pi/2);
arclength(xint,ffun(xint),'spline')
ans =
     4.7383      

which also seems to agree pretty well.

[Opariuc Andrei]

thx :)